Compute $P(X<Y).$

Question

Let $X$ and $Y$ be two independent random variables with pdf's $f_X(x)=1, \ 0<x<1$ and $f_Y(x)=3x^2, \ 0<x<1.$ Compute $P(X<Y).$

Since $X,Y$ are independet their joint pdf is $f(x,y)=1\cdot 3x^2=3x^2, \ 0<x,y<1.$ So,

$$P(X<Y)=3\int_0^1\int_0^yx^2 \ dxdy=\frac{1}{4}.$$

My the exam solutions say the answer is $3/4$. Who is correct?

Answer 1

The exam solution is correct.

If $[x<y]$ denotes the function $\mathbb R^2\to\mathbb R$ taking value $1$ if $x<y$ and taking value $0$ otherwise then:$$P(X<Y)=\int_0^1\int_0^1[x<y]f_{X,Y}(x,y)dxdy=\int_0^1\int_0^1[x<y]f_{X}(x)f_{Y}(y)dxdy=$$$$\int_0^1\int_0^1[x<y]3y^2dxdy=3\int_0^1\int_0^yy^2dxdy=3\int_0^1y^2\int_0^ydxdy=3\int_0^1y^3dy=\frac34$$

Answer 2

Without doing any calculations, I'd first guess that your book is correct: roughly, $Y$ is more dense around $1$, and is 'more likely' to be large. So, intuitively we ought to expect $Y$ is likely to be larger than $X$, which is just uniform.

Your problem was in making the joint pdf: Remember what the variables $x$ and $y$ mean in context. If $X$ had pdf $f_X(x) = \frac{1}{2}x^2$ and $Y$ had pdf $f_Y(x) = \frac{1}{3}x^3$ (independently), you wouldn't say the joint pdf is $f_{X,Y}(x,y) = \frac{1}{6} x^5$ - you'd relabel $f_Y(y) = \frac{1}{3}y^3$ and correctly write $f_{X,Y}(x,y) = \frac{1}{6}x^2 y^3$.

You've made a similar error here, you should get $f(x,y) = 3y^2$.